## College Physics (4th Edition)

Published by McGraw-Hill Education

# Chapter 6 - Problems - Page 229: 33

#### Answer

(a) The cart is moving with a speed of $14.3~m/s$ at point 3. (b) The cart will reach point 4.

#### Work Step by Step

(a) We can use the conservation of energy to find the speed of the cart at point 3. WE can let $U_1 = 0$: $KE_3+U_3 = KE_1+U_1$ $KE_3 = KE_1+0-U_3$ $\frac{1}{2}mv_3^2 = \frac{1}{2}mv_1^2-mgh_3$ $v_3^2 = v_1^2-2gh_3$ $v_3 = \sqrt{v_1^2-2gh_3}$ $v_3 = \sqrt{(20.0~m/s)^2-(2)(9.81~m/s^2)(10.0~m)}$ $v_3 = 14.3~m/s$ The cart is moving with a speed of $14.3~m/s$ at point 3. (b) Let's assume that all the initial kinetic energy is converted to gravitational potential energy. We can find the maximum height the cart can reach: $U_g = KE_1$ $mgh = \frac{1}{2}mv^2$ $h = \frac{v^2}{2g}$ $h = \frac{(20.0~m/s)^2}{(2)(9.81~m/s^2)}$ $h = 20.4~m$ Since the cart can reach a maximum height of 20.4 meters, the cart will reach point 4 which is at a height of 20.0 meters.

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